It suffices to show existence, because the convexity of the functional \(\mathcal {F}_{\chi }(\mathbf {f})\)  implies uniqueness. Let \(\lbrace \mathbf {f}^{(n)}\rbrace _{n=1}^\infty \)  be a minimizing sequence for \(\mathcal {F}_{\chi ;\textup {min}}(M)\) . Then, passing to a subsequence, we may assume that for each \(j\) , \({f}_j^{(n)}\rightharpoonup f_j^*\)  in \(L^r(\omega \times \mathbb {R}^2)\)  for any \(1<r<\infty \)  so that \(\mathbf {f}^*\in \mathcal {A}_{\textup {c.m.}}\) . We will that \(\mathbf {f}^*\)  is a minimizer for \(\mathcal {F}_{\chi ;\textup {min}}(M)\) .
Note that by [1]} and the positive terms in the series, the following quantities are reduced in the weak limit;
\(\liminf _{n\rightarrow \infty }\sum _{j=1}^\infty \Vert |v|^2 {f}_j^{(n)}\Vert _{L^1(\omega \times \mathbb {R}^2)}\ge \sum _{j=1}^\infty \Vert |v|^2 {f}_j\Vert _{L^1(\omega \times \mathbb {R}^2)},\quad \liminf _{n\rightarrow \infty }\mathcal {C}_\beta (\mathbf {f}^{(n)}) \ge \mathcal {C}_\beta (\mathbf {f}^*)\) 
and
\(\begin{aligned}&\liminf _{n\rightarrow \infty }\sum _{j=1}^\infty \iint _{\omega \times \mathbb {R}^2} \langle (-\tfrac{\partial _z^2}{2}+V_{\textup {ext}}) \chi _j, \chi _j\rangle _{L^2 (0,1)}{f}_j^{(n)}dydv\\&\ge \sum _{j=1}^\infty \iint _{\omega \times \mathbb {R}^2}\langle (-\tfrac{\partial _z^2}{2}+V_{\textup {ext}}) \chi _j, \chi _j\rangle _{L^2 (0,1)}{f}_j^* dydv.\end{aligned}\) 
In addition, by the uniform weighted norm bound (Lemma REF ), we have
\(\mathcal {F}_{\chi ; \textup {min}}(M)\gtrsim \liminf _{n\rightarrow \infty }\sum _{j=1}^\infty j^2\Vert f_j^{(n)}\Vert _{ L^1(\omega \times \mathbb {R}^2)}\ge \sum _{j=1}^\infty j^2\Vert f_j^*\Vert _{ L^1(\omega \times \mathbb {R}^2)}.\) 
Thus, except the nonlinear potential energy, all terms in the free energy are clearly reduced in the weak limit (see (REF ) and (REF )). For the nonlinear potential energy, we claim that as \(n\rightarrow \infty \) ,
\(\rho _{(\mathbf {f}^{(n)}, \chi )}\rightharpoonup \rho _{(\mathbf {f}^*, \chi )} \textup { in } L^\frac{6}{5}(\Omega ).\) 
Indeed, by the density estimate (Lemma REF ), it suffices to show that for \(g \in C_c^\infty (\omega )\) ,
\(\lim _{n\rightarrow \infty }\int _\Omega (\rho _{(\mathbf {f}^{(n)}, \chi )}- \rho _{(\mathbf {f}^*, \chi )})gdy=0.\) 
For small \(\epsilon >0\) , by the weighted norm bound (Lemma REF ), taking large \(J_\epsilon \ge 1\) , we obtain
\(\int _\Omega (\rho _{(\mathbf {f}^{(n)}, \chi )}-\rho _{(\mathbf {f}^*, \chi )})g dx=\sum _{j=1}^{J_\epsilon }\int _\Omega (\rho _{f_j^{(n)}}-\rho _{f_j^*})|\chi _j|^2g dx+O(\epsilon ).\) 
On the other hand, we have
\(\int _\Omega (\rho _{f_j^{(n)}}-\rho _{f_j^*})|\chi _j|^2g dx=\int _{\omega \times \mathbb {R}^2} (f_j^{(n)}-f_j^*)\mathbb {1}_{|v|\le R}\left(\int _0^1|\chi _j|^2gdz\right) dx+O(R^{-2})\rightarrow 0\) 
taking \(n\rightarrow \infty \)  and then \(R\rightarrow \infty \) . Hence, the claim (REF ) is proved. Now, we observe from Lemma REF  that the functional \(\rho \mapsto \int _\Omega |\nabla U_\rho |^2dx\)  is lower semi-continuous on \(L^\frac{6}{5} (\Omega )\rightarrow \mathbb {R}\)  in the strong topology. Thus, by the weak convergence (REF ) and the convexity of the map \(\rho \mapsto \int _\Omega |\nabla U_\rho |^2dx\) , we conclude that
\(\liminf _{n\rightarrow \infty }\Vert \nabla U_{(\mathbf {f}^{(n)},\chi )}\Vert _{L^2(\Omega )}\ge \Vert \nabla U_{(\mathbf {f}^*,\chi )}\Vert _{L^2(\Omega )}\) 
holds (see [1]}). Therefore, we conclude that
\(\mathcal {F}_{\chi ;\textup {min}}(M)=\liminf _{n\rightarrow \infty }\mathcal {E}(\mathbf {f}^{(n)},\chi )\ge \mathcal {E}(\mathbf {f},\chi ).\) 
It remains to show that \(\mathbf {f}^*=\lbrace f_j^*\rbrace _{j=1}^\infty \)  is admissible for the problem \(\mathcal {F}_{\chi ;\textup {min}}(M)\) . Indeed, for any small \(\epsilon >0\) , by Lemma REF  and (REF ), there exist large \(J_\epsilon , R_\epsilon \ge 1\)  such that
\(\begin{aligned}\mathcal {M}(\mathbf {f}^{(n)})&=\sum _{j=1}^{J_\epsilon } \iint _{\omega \times \mathbb {R}^2}\mathbb {1}_{|v|\le R_\epsilon }{f}_j^{(n)}dydv+O(\epsilon ),\\\mathcal {M}(\mathbf {f}^*)&=\sum _{j=1}^{J_\epsilon } \iint _{\omega \times \mathbb {R}^2}\mathbb {1}_{|v|\le R_\epsilon }{f}_j^*dydv+O(\epsilon ).\end{aligned}\) 
Hence, taking \(n\rightarrow \infty \) , it follows from that
\(M=\mathcal {M}(\mathbf {f}^{(n)})\rightarrow \sum _{j=1}^{J_\epsilon } \iint _{\omega \times \mathbb {R}^2} \mathbb {1}_{|v|\le R_\epsilon }{f}_j^*dydv+O(\epsilon )=\mathcal {M}(\mathbf {f}^*)+O(\epsilon ).\) 
This proves that \(\mathcal {M}(\mathbf {f}^*)=M\) , since \(\epsilon >0\)  is arbitrary.
